What is b if f is continuous at x=π/2?

b=8/3. Using limit substitution h=π/2−x: b(1−cos h)/(4h²)→b/8=1/3, giving b=8/3.

What is ∫₀²|x²+2x−3|dx?

4. Split at x=1 (root in [0,2]): ∫₀¹(3−x²−2x)dx+∫₁²(x²+2x−3)dx=5/3+7/3=4.

Piecewise f: 1/3 for x≤π/2 and b(1−sinx)/(π−2x)² for x>π/2, continuous at π/2 — find ∫₀^(3b−6)|x²+2x−3|dx

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MathCalculus

QMCQCalculus

Let $f(x)=\begin{cases}\dfrac{1}{3},&x\leq\dfrac{\pi}{2}\\[10pt]\dfrac{b(1-\sin x)}{(\pi-2x)^2},&x>\dfrac{\pi}{2}\end{cases}$. If $f$ is continuous at $x=\dfrac{\pi}{2}$, then the value of $\displaystyle\int_0^{3b-6}\bigl|x^2+2x-3\bigr|\,dx$ is:

A) $5$ B) $2$ C) $3$ D) $4$

✅ Correct Answer

D) 4

✓

Solution

Find b using continuity at x = π/2

We need $\displaystyle\lim_{x\to(\pi/2)^+}\dfrac{b(1-\sin x)}{(\pi-2x)^2}=f\!\left(\dfrac{\pi}{2}\right)=\dfrac{1}{3}$.

Let $x=\dfrac{\pi}{2}-h$ where $h\to0^+$, so $\pi-2x=2h$ and $\sin x=\cos h$:

$\displaystyle\lim_{h\to0^+}\frac{b(1-\cos h)}{4h^2}=\frac{b}{4}\cdot\lim_{h\to0^+}\frac{1-\cos h}{h^2}=\frac{b}{4}\cdot\frac{1}{2}=\frac{b}{8}$

$\therefore\quad\dfrac{b}{8}=\dfrac{1}{3}\quad\Rightarrow\quad b=\dfrac{8}{3}$

Find the upper limit 3b − 6

$3b-6=3\cdot\dfrac{8}{3}-6=8-6=2$

Evaluate ∫₀² |x²+2x−3| dx

$x^2+2x-3=(x+3)(x-1)$. In $[0,2]$: root at $x=1$ (since $x=-3\notin[0,2]$).

For $x\in[0,1]$: $x^2+2x-3<0$, so $|x^2+2x-3|=3-x^2-2x$.

For $x\in[1,2]$: $x^2+2x-3>0$, so $|x^2+2x-3|=x^2+2x-3$.

$I=\displaystyle\int_0^1(3-x^2-2x)\,dx+\int_1^2(x^2+2x-3)\,dx$

$=\left[3x-\dfrac{x^3}{3}-x^2\right]_0^1+\left[\dfrac{x^3}{3}+x^2-3x\right]_1^2$

$=\left(3-\dfrac{1}{3}-1\right)+\left(\dfrac{8}{3}+4-6-\dfrac{1}{3}-1+3\right)$

$=\dfrac{5}{3}+\left(\dfrac{7}{3}\right)=\dfrac{12}{3}=\boxed{4}$

📘 Continuity and Absolute Value Integration

For continuity: equate the limit from the discontinuous side to the known value. The limit $\lim_{h\to0}(1-\cos h)/h^2=1/2$ (standard result) is essential. For integrals with $|f(x)|$: find roots of $f(x)$ in the domain, split the integral, and integrate without absolute value signs (with appropriate sign changes).

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Important Concepts

Standard Limit Used$\lim_{h\to0}\dfrac{1-\cos h}{h^2}=\dfrac{1}{2}$. Proof: using $1-\cos h=2\sin^2(h/2)$ and $\lim_{\theta\to0}(\sin\theta/\theta)=1$: $\dfrac{2\sin^2(h/2)}{h^2}=\dfrac{\sin^2(h/2)}{2(h/2)^2}\to\dfrac{1}{2}$.

Substitution h = π/2 − xSetting $x=\pi/2-h$ converts $1-\sin x=1-\cos h$ and $\pi-2x=2h$. This makes the limit a standard form.

Splitting |x²+2x−3|Root of $x^2+2x-3=(x+3)(x-1)$ in $[0,2]$ is $x=1$. Left of 1: polynomial is negative (e.g., at $x=0$: $0+0-3=-3<0$). Right of 1: positive. Split accordingly.

Verification of AnswerPart 1: $\int_0^1(3-x^2-2x)dx=[3x-x^3/3-x^2]_0^1=3-1/3-1=5/3$. Part 2: $\int_1^2(x^2+2x-3)dx=[x^3/3+x^2-3x]_1^2=(8/3+4-6)-(1/3+1-3)=2/3-(-5/3)=7/3$. Total $=5/3+7/3=4$ ✓

FAQs

How does substitution h=π/2−x convert the limit?

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When $x>\pi/2$: $x=\pi/2+h$ as $h\to0^+$. Then $\sin x=\sin(\pi/2+h)=\cos h$, so $1-\sin x=1-\cos h$. Also $\pi-2x=\pi-2(\pi/2+h)=-2h$, so $(\pi-2x)^2=4h^2$.

What is lim(1−cos h)/h² as h→0?

⌄

$(1-\cos h)/h^2=2\sin^2(h/2)/h^2=\frac{1}{2}\cdot(\sin(h/2)/(h/2))^2\to 1/2$.

Why does 3b−6=2?

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$b=8/3$: $3\cdot(8/3)-6=8-6=2$.

What is the sign of x²+2x−3 on [0,1]?

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At $x=0$: $0+0-3=-3<0$. At $x=1$: $1+2-3=0$. So negative throughout $[0,1)$.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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